The Fourier-Motzkin elimination (FME) is a procedure that reduces an $n$-variable problem to an equivalent $(n − 1)$-variable problem. It is analogous to Gaussian elimination but for a system of inequalities. In this post, it was shown why Gaussian elimination does not change the solution set. Does Fourier-Motzkin elimination also not change the solution set? Why or why not?
2026-02-23 10:16:45.1771841805
Does Fourier-Motzkin elimination change the solution set?
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I think the answer is affirmative. By what I know of the Fourier-Motzkin elimination: FME first classifies the inequalities into three kinds and then change the coefficient of the first variable $x_1$ to $1$ or 0, this first step does not change the solution set apparently; then FME kicks $x_1$ out of the game by observing that all possible values of $x_1$ is already obtained for a fixed $x' = (x_2,...,x_n)$, this doesn't change the solution set too. Hence, in conclusion, Fourier-Motzkin elimination also not change the solution set.